What is Henry's law constant of a gas if solubility of gas in water at 25^0C is 0.028 mol ,dm^ -3 [partial pressure of gas = 0.346 bar]

MHT CET 2023 Chemistry question

What is Henry's law constant of a gas if solubility of gas in water at C is 0.028 [partial pressure of gas = 0.346 bar]

(A)0.081 Correct
(B)0.075
(C)0.093
(D)0.049

Answer: (A)

Solution

Henry's law in the form , where is the solubility (mol dm) and is the partial pressure of the gas. So K_H = \dfrac{S}{p} = \dfrac{0.028\ \mathrm{mol\,dm^{-3}}}{0.346\ bar} = 0.0809\ \mathrm{mol\,dm^{-3}\,bar^{-1}} \approx 0.081,

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Question ID VK-098197